1.

A square loop of side 10 cm and resistance 0.5 Omega is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.

Answer»

Solution :The angle `theta`made by the area vector of the coil with the magnetic FIELD is `45^@` . From EQ. (6.1), the initial magnetic flux is
`Phi = BA cos theta`
`=(0.1xx10^(-2))/(sqrt2)WB`
Final flux, `Phi_(min)=0`
The CHANGE in flux is brought about in 0.70 s. From Eq. (6.3), the magnitude of the INDUCED emf is given by
`epsi =(|DeltaPhi_(B)|)/(Deltat) = (|Phi-0|)/(Deltat) = (10^(-3))/(sqrt2 xx 0.7) = 1.0 m V`
And the magnitude of the current is
`I = epsi/R = (10^(-3)V)/(0.5 Omega) = 2mA`
Note that the earth’s magnetic field also produces a flux through the loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf.


Discussion

No Comment Found

Related InterviewSolutions